If Hund rule violate, then find the total number of species among following which whill be dimagnetic:
`B_(2), O_(2), N_(2)^(-), C_(2), NO, OF, N_(2)^(2-), BN`

Answer

To determine the total number of diamagnetic species from the given list, we will analyze each molecule based on its electron configuration and the presence of unpaired electrons. A species is considered diamagnetic if all its electrons are paired.

### Step-by-Step Solution:

1. **Identify the total number of electrons for each species:**
- \( B_2 \): Each boron atom has 5 electrons, so \( B_2 \) has \( 5 + 5 = 10 \) electrons.
- \( O_2 \): Each oxygen atom has 8 electrons, so \( O_2 \) has \( 8 + 8 = 16 \) electrons.
- \( N_2^- \): Each nitrogen atom has 7 electrons, and with a -1 charge, it has \( 7 + 7 + 1 = 15 \) electrons.
- \( C_2 \): Each carbon atom has 6 electrons, so \( C_2 \) has \( 6 + 6 = 12 \) electrons.
- \( NO \): Nitrogen has 7 electrons and oxygen has 8, so \( NO \) has \( 7 + 8 = 15 \) electrons.
- \( OF \): Oxygen has 8 electrons and fluorine has 9, so \( OF \) has \( 8 + 9 = 17 \) electrons.
- \( N_2^{2-} \): Each nitrogen has 7 electrons, and with a -2 charge, it has \( 7 + 7 + 2 = 16 \) electrons.
- \( BN \): Boron has 5 electrons and nitrogen has 7, so \( BN \) has \( 5 + 7 = 12 \) electrons.

2. **Determine the electron configuration for each species:**
- **For \( B_2 \) (10 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- All electrons are paired → **Diamagnetic**.

- **For \( O_2 \) (16 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \)
- Contains unpaired electrons → **Paramagnetic**.

- **For \( N_2^- \) (15 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \)
- Contains unpaired electrons → **Paramagnetic**.

- **For \( C_2 \) (12 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- All electrons are paired → **Diamagnetic**.

- **For \( NO \) (15 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \)
- Contains unpaired electrons → **Paramagnetic**.

- **For \( OF \) (17 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \)
- Contains unpaired electrons → **Paramagnetic**.

- **For \( N_2^{2-} \) (16 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- All electrons are paired → **Diamagnetic**.

- **For \( BN \) (12 electrons):**
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- All electrons are paired → **Diamagnetic**.

3. **Count the total number of diamagnetic species:**
- From the analysis, the diamagnetic species are \( B_2 \), \( C_2 \), \( N_2^{2-} \), and \( BN \).
- Total = 4 diamagnetic species.

### Final Answer:
The total number of diamagnetic species among the given options is **4**.